Order = highest derivative. Degree = power of highest derivative. When degree is undefined.

Order and Degree of a Differential Equation — How to Find

Question

Find the order and degree of the following differential equations:

$\dfrac{d^2y}{dx^2} + 3\left(\dfrac{dy}{dx}\right)^3 - 5y = 0$
$\left(\dfrac{d^3y}{dx^3}\right)^2 + \sin\left(\dfrac{dy}{dx}\right) = 0$
$\dfrac{dy}{dx} = \sqrt{1 + \left(\dfrac{d^2y}{dx^2}\right)^2}$

Solution — Step by Step

Order is the highest-order derivative present in the equation. Degree is the power of that highest-order derivative — but only after the equation is made polynomial in derivatives (no roots, no trig, no logs wrapping the derivatives).

The derivatives present are $\dfrac{d^2y}{dx^2}$ and $\dfrac{dy}{dx}$ . The highest is the second derivative, so order = 2.

Now look at the power on that second derivative: $\left(\dfrac{d^2y}{dx^2}\right)^1$ — the power is 1. So degree = 1.

The highest derivative is $\dfrac{d^3y}{dx^3}$ , so order = 3. But look at the second term: $\sin\left(\dfrac{dy}{dx}\right)$ .

We cannot express $\sin\left(\dfrac{dy}{dx}\right)$ as a finite polynomial in $\dfrac{dy}{dx}$ — expanding sine gives an infinite series. The equation is not polynomial in its derivatives, so degree = undefined (not defined).

The equation has $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ wrapped under a root. We must eliminate the radical before assigning a degree.

Square both sides:

\left(\dfrac{dy}{dx}\right)^2 = 1 + \left(\dfrac{d^2y}{dx^2}\right)^2

Now the equation is polynomial in derivatives. Highest derivative is $\dfrac{d^2y}{dx^2}$ , with power 2. So order = 2, degree = 2.

Equation	Order	Degree
Equation 1	2	1
Equation 2	3	Not defined
Equation 3	2	2

Why This Works

Order is easy — just spot which derivative goes the deepest. $\dfrac{d^3y}{dx^3}$ beats $\dfrac{d^2y}{dx^2}$ beats $\dfrac{dy}{dx}$ .

Degree requires one extra condition: the equation must be expressible as a polynomial in its derivatives. Think of it like degree of a polynomial in $x$ — if $x$ is stuck inside $\sin x$ or $e^x$ , we can’t assign a polynomial degree. Same logic applies here with the derivatives.

The square root case catches students off guard. We always squash roots (or any fractional power) by raising both sides to the appropriate power before reading off the degree.

Alternative Method

Shortcut for spotting “degree not defined”: If any derivative appears inside $\sin$ , $\cos$ , $\log$ , $e^{(\cdot)}$ , or under a root that cannot be cleared with a single squaring, the degree is immediately not defined. You don’t need to expand anything — just check whether derivatives are “trapped” inside transcendental functions.

For Equation 3, some students prefer to rewrite and check directly:

\dfrac{dy}{dx} = \sqrt{1 + \left(\dfrac{d^2y}{dx^2}\right)^2}

The root wraps the second derivative, not the first. One squaring clears it completely:

\left(\dfrac{dy}{dx}\right)^2 - 1 = \left(\dfrac{d^2y}{dx^2}\right)^2

Highest derivative is $\dfrac{d^2y}{dx^2}$ , raised to power 2. Order = 2, degree = 2. Same answer.

Common Mistake

Confusing the order with the degree. In Equation 1 — $\dfrac{d^2y}{dx^2} + 3\left(\dfrac{dy}{dx}\right)^3 - 5y = 0$ — many students write degree = 3 because they see the $(\cdot)^3$ on the first derivative. Wrong. Degree is the power on the highest derivative, which is $\dfrac{d^2y}{dx^2}$ here — and its power is 1.

The $\left(\dfrac{dy}{dx}\right)^3$ is tempting, but it belongs to a lower-order derivative. It has no role in determining the degree.

This exact confusion appears every year in CBSE Class 12 board exams. The question is easy marks — 1–2 marks typically — so nail the definition: degree always refers to the power on the highest-order derivative after clearing any radicals or fractional powers.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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